Electrical and electronics measurements miscellaneous Difficult Questions and Answers | Page - 5

Electrical and electronics measurements miscellaneous

A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type ac voltmeter connected in parallel. If the meter readings are V₁ and V₂ respectively and the meters are free from calibration errors, then the form factor of the ac voltage may be estimated as

1. V₁
  V₂
1. 1.11V₁
  V₂
1. √2V₁
  V₂
1. πV₁
  V₂

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Form factor of the wave = rms Value
Mean Value

Moving iron Instrument will show rms value. Rectifier voltmeter is calibrate to read rms value of sinusoidal voltage, i.e. with form factor of 1.11.
∴ Mean value of the applied voltage = V₂
1.11

∴ Form factor = V₁ = 1011V₁
V₂/1.11 V₂

Correct Option: B

Form factor of the wave = rms Value
Mean Value

Moving iron Instrument will show rms value. Rectifier voltmeter is calibrate to read rms value of sinusoidal voltage, i.e. with form factor of 1.11.
∴ Mean value of the applied voltage = V₂
1.11

∴ Form factor = V₁ = 1011V₁
V₂/1.11 V₂

When damping of an instrument is adjusted to enable the pointer to rise quickly to its deflected position without overshooting, in that case the instrument is said to be

1. dead-beat
1. off-beat
1. over-damped
1. under-damped

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NA

Correct Option: A

NA

The resistance of two coils of a wattmeter are 0-0.1 ohm and 1000 ohms respectively and both are non-inductive. The load current is 20 A and the voltage across the load is 30 V. In one of the two ways of connecting the voltage coil, the error in the reading would be

1. 0.1% too high
1. 0.2% too high
1. 0.15% too high
1. zero

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r_cc = 0.01 ohm, r_vc = 1000 ohm.
There are two ways in which the current coil and voltage coil can be connected. The wattmeter reads high by an amount equal to power loss in current coil in one case and in voltage coil in the other. The two possible power losses are
(i) I_C² r_cc = 20² × 0.01 = 4W
(ii)
V² = 30² = 0.9 W
r_cc 1000

The two percentage errors are
(i)
4 × 100 = 2
600 3

(ii)
0.9 × 100 = 0.15
600

Correct Option: C

r_cc = 0.01 ohm, r_vc = 1000 ohm.
There are two ways in which the current coil and voltage coil can be connected. The wattmeter reads high by an amount equal to power loss in current coil in one case and in voltage coil in the other. The two possible power losses are
(i) I_C² r_cc = 20² × 0.01 = 4W
(ii)
V² = 30² = 0.9 W
r_cc 1000

The two percentage errors are
(i)
4 × 100 = 2
600 3

(ii)
0.9 × 100 = 0.15
600

The measurement of a quantity

1. is an act of comparison of an unknown quantity with another quantity
1. is a act of comparison of an unknown quantity with a known quantity whose accuracy may be known or may not be known
1. is an act of comparison of an unknown quantity with a predefined acceptable standard which is accurately known
1. none of these

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NA

Correct Option: C

NA

A piezo electric transducer has the following parameter values:
Crystal capacitance = 10^–9 F
Cable capacitance = 2 × 10^–10 F
Charge sensitivity = 4 × 10^–6 coulomb/cm
If the oscilloscope used for read-out has an input resistance of 1 MW in parallel with C = 4 × 10^–10 F, then the voltage sensitivity constant will be

1. 2500 V/cm
1. 3334 V/cm
1. 4000 V/cm
1. 4500 V/cm

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Total capacitance = Crystal capacitance + Cable capacitance + Oscilloscope capacitance
= (10 + 2 + 4) × 10^–10 F
Change sensitivity = 4 × 10⁶ C/cm
Voltage sensitivity = Change sensitivity
Total capacitance

= 4 × 10⁶ = 2500 V/cm
16 × 10^–10

Correct Option: A

Total capacitance = Crystal capacitance + Cable capacitance + Oscilloscope capacitance
= (10 + 2 + 4) × 10^–10 F
Change sensitivity = 4 × 10⁶ C/cm
Voltage sensitivity = Change sensitivity
Total capacitance

= 4 × 10⁶ = 2500 V/cm
16 × 10^–10